HDU 4424 - Conquer a New Region(最大生成树)

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题目:

http://acm.hdu.edu.cn/showproblem.php?pid=4424

题意:

n个点的树,i点到j点的路径最大价值是i到j点的路的权值最小值,以一个点为源点,使得从这个点到其他点的总价值最大。

思路:

将边权值按照从大到小排序,枚举每一条边,此时的边为当前所有边的最小边,判断将u集合合并到v集合还是 v集合合并到u集合,使用并查集实现。最后集合的祖先就是源点。

AC.

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int maxn = 2e5+5;struct Edge{    int u, v, w;    bool operator < (const Edge &A) const{        return w > A.w;    }}edge[maxn];int fa[maxn];int find(int x){    if(fa[x] == x) return x;    else return fa[x] = find(fa[x]);}int num[maxn];void unite(int x, int y){    x = find(x), y = find(y);    if(x != y) {        fa[y] = x;        num[x] += num[y];    }}long long sum[maxn];void init(int n){    for(int i = 1; i <= n; ++i) {        fa[i] = i;        num[i] = 1;    }    memset(sum, 0, sizeof(sum));}int main(){    //freopen("in", "r", stdin);    int n;    while(~scanf("%d", &n)) {        init(n);        for(int i = 0; i < n-1; ++i) {            scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].w);        }        sort(edge, edge+n-1);        long long s1, s2;        for(int i = 0; i < n-1; ++i) {            int u = edge[i].u, v = edge[i].v, w = edge[i].w;            int fu = find(u), fv = find(v);            if(fu != fv) {                s1 = (long long)w*num[fu] + sum[fv];                s2 = (long long)w*num[fv] + sum[fu];                if(s1 > s2) {                    unite(fv, fu);                    sum[fv] = s1;                }                else {                    unite(fu, fv);                    sum[fu] = s2;                }            }        }        printf("%I64d\n", sum[find(1)]);    }    return 0;}


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